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A conductor of length $l$ is placed perpendicular to a horizontal uniform magnetic field. Suddenly, a certain amount of charge is passed through it, when it is found to jump to a height $h$ . The amount of charge that passes through the conductor is

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$\frac{m \sqrt{g h}}{B l}$

$\frac{m \sqrt{g h}}{2 B l}$

Noen of these

$\frac{m \sqrt{2 g h}}{B l}$

answer is C.

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Detailed Solution

Linear impulse $= m v$

or, $F ∆ t = m \sqrt{2 g h}$

or, $(i l B) ∆ t = m \sqrt{2 g h}$

But, $i ∆ t = ∆ q$

$∴$ $(∆ q) (l b) = m \sqrt{2 g h}$

Hence, $∆ q = \frac{m \sqrt{2 g h}}{B l}$

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